How does one determine the yield point of a wooden or aluminum mast?

[laukejas]

Too bad you can't get a copy of 'Skene's Element of Yacht Design" edited by Francis Kinney. pg 163 - 190 or Sailing Yacht Design by Henry & Miller. Each has some good info & calcs on wooden and aluminum spars. There may be some SNAME papers as well.
As to yield point, not sure about wood but aluminum can be from suppliers.

I actually do have "Skene's Element of Yacht Design", 2001 edition. My very good friend from Florida donated this book. However, I do not find anything about this in pages 163-190. There is the formula about sizing wooden masts which I quoted in my first post on this topic. I'm still looking through the book now.

Skene does give a very handy formula P= π^2*EL / L^2, which works both for aluminum and wooden spars, but it tells breaking load for the compressive load, not bending. Damn. If only I found a similar formula for bending loads...

P.S. Wait!!! I think I found it on page 27!!! Reading it now

 

[Ad Hoc]

I am an amateur in engineering field, so this is where my knowledge ends. I know that each material has it's Yield Point, past which it will deform permanently, and break soon after....Wood and aluminum probably also have a very different Yield Points. For example, I can calculate that my 100" mast will bend 6 3/4" in specific load conditions. But will it recover from such bend? Or will it stay deformed?

You have answered your own question

All structures are designed to satisfy both stresses and defection criterion that is set as 'acceptable'. In the case of stress, the maximum stress is the yield stress. Since all loading and subsequent unloading of the structure/material if below the yield point is linear and as such the structure/material shall return to its unloaded non-deformed shape.

The difficulty is knowing if the loads you are applying and consistent and constant. If not then factors of safety are added to the applied load to ensure that when loaded, the structure/material never exceeds the yield point. Because if it does, then you have problems you are noting..permanent deformation and/or catastrophic failure.

 

[laukejas]

You have answered your own question

All structures are designed to satisfy both stresses and defection criterion that is set as 'acceptable'. In the case of stress, the maximum stress is the yield stress. Since all loading and subsequent unloading of the structure/material if below the yield point is linear and as such the structure/material shall return to its unloaded non-deformed shape.

The difficulty is knowing if the loads you are applying and consistent and constant. If not then factors of safety are added to the applied load to ensure that when loaded, the structure/material never exceeds the yield point. Because if it does, then you have problems you are noting..permanent deformation and/or catastrophic failure.

No, I haven't answered my question, because I still do not know if that 6 3/4" deflection is below or above the Yield Point... I don't know how to determine it.

There is something in Skene's book about Breaking Load for cantilever beam load. I'm trying to figure it out. This might be it.

 

[Ad Hoc]

No, I haven't answered my question, because I still do not know if that 6 3/4" deflection is below or above the Yield Point... I don't know how to determine it.

Ok..baby steps.

How did you calculate the deflection?

 

[laukejas]

Ok..baby steps.

How did you calculate the deflection?

Well, θ = Pa^2 / 2EI.

Where θ is deflection, P is is the force, a is center of effort above mast partner, E is Elastic Modulus of spruce, and I is Moment of Inertia.

 

[Ad Hoc]

So if you know E...you know what material you are using. Thus you should know what is the yield stress of THAT material

 

[Heimfried]

[...] Center of effort is, say, 50" above mast partner, and the luff is attached, so using the Cantilever Beam deflection formula for concentrated load P at any point (it's not really concentrated, but let's do concentrated for simplicity), we get:
θ = Pa^2 / 2EI, where P is is the force, a is center of effort above mast partner, E is Elastic Modulus of spruce (9 GPa), and I is Moment of Inertia. Once you plug in the numbers, the result is 6 3/4" of deflection at the mast top. Sounds about right. [...]

Little mistake: the result of Pa^2 / 2 EI is not the deflection (lenght), but the angle (therefor it is called theta or phi).
The deflection is w = Pa^3 / 3 EI.

There is a bending stress sigma = M / W = P * a / W

W is Widerstandsmoment (Section Modulus)

This sigma is to compare with the yield strength.

 

[laukejas]

So if you know E...you know what material you are using. Thus you should know what is the yield stress of THAT material

Well, yes. But what kind of formula do I need to use to translate that yield stress (~5000 MPa for pine) into bending load? In other words, what kind of bending load force would produce those 5000 MPa?

Sorry if I'm missing out something very basic here. I barely finished my maths and physics courses back in high school, and never did anything with it after. I have a lot of catching up to do... Thank you for your patience.

 

[Ad Hoc]

Ahh//H beat me to it..

OK...so now, you have the formula for the bending. So apply the load and check what stress you get compared to the yield stress of the material you are using. Simple

 

[TANSL]

I suppose we all know that by applying a certain bending moment to a beam (the mast) not all points of that beam are subjected to the same stress, nor all points of that beam have the same flexural strength. So, at what point do you want to calculate the yield stress?

 

[UpOnStands]

mast testing to destruction

British style

https://www.youtube.com/watch?v=d-jqwkNXMIY&t=378s

 

[laukejas]

I suppose we all know that by applying a certain bending moment to a beam (the mast) not all points of that beam are subjected to the same stress, nor all points of that beam have the same flexural strength. So, at what point do you want to calculate the yield stress?

At the weakest point, of course! Wherever it is.

Little mistake: the result of Pa^2 / 2 EI is not the deflection (lenght), but the angle (therefor it is called theta or phi).
The deflection is w = Pa^3 / 3 EI.

Thanks for the correction! I overlooked that one.

There is a bending stress sigma = M / W = P * a / W

W is Widerstandsmoment (Section Modulus)

This sigma is to compare with the yield strength.

Now this, THIS is what I was looking for! Thank you, kind sir. I must admit, I expected the formula to be far more complex than this. I haven't tested it yet, but it looks right. Ad Hoc, thank you very much too.

Incidentally, I found a little different equation to calculate breaking load in Skene's "Elements of Yacht Design", page 27:



As you can see, his formula uses Moment of Inertia of Section (I), rather than Section Modulus (S). But since S is equal to I / f, I guess it's the same thing.

I assume the max. fiber stress given in the table is the Ultimate Tensile Strength. So I guess the same formula can be used to determine the load that would deform the mast by comparing the calculated load with Yield Strength instead. Correct?

 

[Heimfried]

[...]
As you can see, his formula uses Moment of Inertia of Section (I), rather than Section Modulus (S). But since S is equal to I / f, I guess it's the same thing. [...]

S is not eaqual to I / f, I think.

E. g. for a circular tube
I = pi / 64 * (D^4 - d^4)
S = pi / 32 * (D^4 - d^4) / D

 

[Heimfried]

Look at this table (I found it only in the German version):

https://de.wikipedia.org/wiki/Streckgrenze#Dehngrenze_.28.3D_Ersatzstreckgrenze.29

For Aluminium are values shown from 40 N/mm^2 to 525 n/mm^2 depending at the specific alloy.

 

[laukejas]

S is not eaqual to I / f, I think.

E. g. for a circular tube
I = pi / 64 * (D^4 - d^4)
S = pi / 32 * (D^4 - d^4) / D

Oh, my mistake. I meant S = I / y, not f. y is the distance from the neutral axis to any given fibre. I misquoted this formula from Wikipedia article on Section Modulus.

Look at this table (I found it only in the German version):

https://de.wikipedia.org/wiki/Streckgrenze#Dehngrenze_.28.3D_Ersatzstreckgrenze.29

For Aluminium are values shown from 40 N/mm^2 to 525 n/mm^2 depending at the specific alloy.

Thanks for the link. I'll use it as a reference. I have trouble finding values for wood. Skene quotes maximum fiber stress as 5000 lbs for spruce. I wonder why he uses lbs instead of psi. Anyway, I can't find any info that would confirm that number, either in metric or imperial. Wiki article in English about ultimate tensile strength gives 40 N/mm^2 for pine, which is around 5800 psi. I wish there was a table for other wood species as well.

Anyway, these are ultimate strength values. What about yield strength values for wood? I suppose yield strength is quite a bit below ultimate strength. Do you know any resource that would quote these numbers?